Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly andeach person arrives at a different time. In how many ways can they arrive? In how many ways can Jean arrivefirst and Keith last? Find the probability that Jean will arrive first and Keith will arrive last.
A) $720 ; 24 ; \frac { 1 } { 30 }$
B) $720 ; 15 ; \frac { 1 } { 48 }$
C) $120 ; 6 ; \frac { 1 } { 20 }$
D) $120 ; 10 ; \frac { 1 } { 12 }$

Question

Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly andeach person arrives at a different time. In how many ways can they arrive? In how many ways can Jean arrivefirst and Keith last? Find the probability that Jean will arrive first and Keith will arrive last. 
A) $720 ; 24 ; \frac { 1 } { 30 }$
B) $720 ; 15 ; \frac { 1 } { 48 }$
C) $120 ; 6 ; \frac { 1 } { 20 }$
D) $120 ; 10 ; \frac { 1 } { 12 }$

Quizplus · Accepted Answer

The total number of ways 6 people can arrive is $6!$ which is $720$. If Jean arrives first and Keith last, there are $4!$ ways for the other 4 to arrive, which is $24$. The probability is $\frac{24}{720} = \frac{1}{30}$.

Amy, Jean, Keith, Tom, Susan, and Dave Have All Been 720;24;130720 ; 24 ; \frac { 1 } { 30 }720;24;301​

Amy, Jean, Keith, Tom, Susan, and Dave Have All Been $720 ; 24 ; \frac { 1 } { 30 }$